The displacement of a damped harmonic oscillator is given by `x(t)=e^(-0.1t)cos(10pit+phi)`. Here t is in seconds
The time taken for its amplitude of vibration to drop to half of its initial is close to :

The displacement of a damped harmonic oscillator is given by x(t)-e^(-0.1t) cos(10pit+varphi) .Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :

The displacement of a simple harmonic oscillator is given by x = 4 cos (2pit+pi//4)m . Then velocity of the oscillator at t = 2 s is

The displacement of a simple harmonic oscillator is given by y= 0.40 sin (440t +0.61) . For this, what are the value of amplitude.

The displacement of a simple harmonic oscillator is x = 5 sin (pi t /3) m . Then its velocity at t =1 s , is

The displacement of a simple harmonic oscillator is, x = 5 sin (pit//3) m . Then its velocity at t = 1 s, is

Equation of motion for a particle performing damped harmonic oscillation is given as x=e^(-0.1t) cos(10pit+phi) . The time when amplitude will half of the initial is :

Equation of motion for a particle performing damped harmonic oscillation is given as x=e^(-1t) cos(10pit+phi) . The time when amplitude will half of the initial is :

Equation of motion for a particle performing damped harmonic oscillation is given as x=e^(-1t) cos(10pit+phi) . The time when amplitude will half of the initial is :

For the damped oscillator, the mass m of the block is 200g, k = 90 Nm^(-1) and the damping constant b is 40 gs^(-1) . Calculate time taken for its amplitude of vibrations to drop to half of its initial value.